Question

which equation or equations represent the vertical asymptote(s) of the function?
$f(x)=\frac{x - 1}{(x + 3)(x - 2)}$
a $x = 3$ and $x = -2$
b $x = 1$
c $x = 1$, $x = -3$, and $x = 2$
d $x = -3$ and $x = 2$

Explanation:

Step1: Recall Vertical Asymptote Rule

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) (and numerator \( N(x)
eq0 \) at those points).

Step2: Find Denominator Zero Points

Given \( f(x)=\frac{x - 1}{(x + 3)(x - 2)} \), set denominator \( (x + 3)(x - 2)=0 \).
Solve \( x + 3 = 0\) gives \( x=-3 \); solve \( x - 2 = 0\) gives \( x = 2 \).
Check numerator at \( x=-3 \) and \( x = 2 \): numerator \( x - 1 \) at \( x=-3 \) is \( -3-1=-4
eq0 \), at \( x = 2 \) is \( 2 - 1 = 1
eq0 \). So vertical asymptotes are at \( x=-3 \) and \( x = 2 \).

Answer:

D. \( x = -3 \) and \( x = 2 \)